Discussion
If \(1, a, b, c, 16\) are in geometric progression, then \(\sqrt[3]{abc}\) is equal to
Answer: D
\(\sqrt[3]{abc}\) equals 4.
With first term 1 and common ratio r, the terms are \(1, r, r^{2}, r^{3}, r^{4}\). So \(r^{4} = 16\) and \(a = r\), \(b = r^{2}\), \(c = r^{3}\):
\[ \begin{aligned} abc &= r^{6} = (r^{2})^{3} \\ \sqrt[3]{abc} &= r^{2} = \sqrt{16} = 4 \end{aligned} \]
This holds whether r is 2 or -2, since only \(r^{2}\) appears. Equivalently, b is the geometric mean of 1 and 16, so b = 4 and \(abc = b^{3}\). 2 is the common ratio, and 8 is the fourth term.
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